2011 AMC 12B Problem 15

Below is the professionally curated solution for Problem 15 of the 2011 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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Concepts:difference of squares prime factorization factor

Difficulty rating: 1740

15.

How many positive two-digit integers are factors of $2^{24}-1?$

$4$

$8$

$10$

$12$

$14$

Solution:

Factoring, $2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^4+1)(2^8-2^4+1),$ which equals $63\cdot65\cdot17\cdot241=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.$

Since $241$ is a three-digit prime, the two-digit factors come from $3^2\cdot5\cdot7\cdot13\cdot17.$ They are $13,15,17,21,35,39,45,51,63,65,85,91,$ for a total of $12.$

Thus, the correct answer is D.

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